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3 years ago

3a+4b=5a-t solve for a

2 answers:

8 0

Solution for 3A-4B+5A= equation:

Simplifying
3A + -4B + 5A = 0

Reorder the terms:
3A + 5A + -4B = 0

Combine like terms: 3A + 5A = 8A
8A + -4B = 0

Solving
8A + -4B = 0

Solving for variable 'A'.

Move all terms containing A to the left, all other terms to the right.

Add '4B' to each side of the equation.
8A + -4B + 4B = 0 + 4B

Combine like terms: -4B + 4B = 0
8A + 0 = 0 + 4B
8A = 0 + 4B
Remove the zero:
8A = 4B

Divide each side by '8'.
A = 0.5B

Simplifying
A = 0.5B

7nadin3 [17]3 years ago

5 0

Answer:

2b+t/2=a

Step-by-step explanation:

3a + 4b = 5a-t

-3a -3a

4b = 2a - t

+t +t

4b + t = 2a

/2 /2

2b+t/2=a

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Answer:

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ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

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Assuming you want the area and perimeter of the shaded part of the shape only...

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Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

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Arc length = $r\theta$ (where r is the radius and $\theta$

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

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Area

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

Perimeter

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

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